91Ó°ÊÓ

Consider the function $f\left(x\right)=2x^2+x-1$.

The function's leading coefficient is positive so the parabola opens upward,

So the ycoordinate of the vertex gives the minimum value of the function.

On comparing the given function $f\left(x\right)=2x^2+x-1$with the standard form $f\left(x\right)=a{x}^2+bx+c$we get

$$\begin{gathered} a=2 \\ b=1 \\ c=-1 \end{gathered}$$

The axis of symmetry for the standard form is $x=-\frac{b}{2a}$.

So for the given function, the axis of symmetry is given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{1}{2×2} \\ x=-0.25 \end{gathered}$$

Thus the equation of the axis of symmetry is $x=-0.25$.

Vertex is a point on the line of axis of symmetry.

Here axis of symmetry is $x=-0.25$.

So the function value $f(-0.25)$is given as

$\begin{array}{rcl}f(-0.25)& =& 2{(-0.25)}^2+(-0.25)-1\\ & =& 0.125-0.25-1\\ & =& -1.125\end{array}$

So the vertex of the parabola is $(-0.25,-1.125)$.

And so the minimum value of the function is$-1.125$.

The graph of the function with its vertex can be seen to verify the result.